Let $p$ be an odd prime integer and let $\zeta$ be a primitive $2p$-th root of unity. Does $\alpha=1+\zeta+\zeta^{-1}+\dots+\zeta^{\frac{p-1}{2}}+\zeta^{-\frac{p-1}{2}}$ generates a normal basis of $Q(\zeta+\zeta^{-1})$. In other words does the Galois conjugates of $\alpha$ form a basis of $\mathbb{Q}(\zeta+\zeta^{-1})$?
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1$\begingroup$ Have you tried writing down some of its conjugates? $\endgroup$– Cam McLemanCommented Apr 6, 2015 at 16:59
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$\begingroup$ Yes! I also have verified that $\alpha$ generates a normal basis for all the odd primes up $p\le 1700$. But I don't have a clue why $\alpha$ always generates a normal basis. $\endgroup$– Angel del RioCommented Apr 6, 2015 at 22:49
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$\begingroup$ I encountered this question when studying the Zassenhaus Conjecture on torsion units of integral group rings of some finite groups. $\endgroup$– Angel del RioCommented Apr 6, 2015 at 22:51
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$\begingroup$ I haven't given this too much thought, but I would've thought the natural approach would've been to write the conjugates in terms of the standard power basis, and then show that the matrix of coefficients was invertible. $\endgroup$– Cam McLemanCommented Apr 7, 2015 at 1:48
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$\begingroup$ Dear Cam, thank you for the hint. I tried to do what you suggest but I don't see any regularity in the matrix of coefficients which can help to obtain . Well, writing $\alpha$ in a power basis (for example, $1$, $\zeta+\zeta^{-1}$, $(\zeta^2+\zeta^{-2}$, $\dots$, seems a bit messy. This is why I prefer to use that more "suitable" basis $b_1=\zeta+\zeta^{-1}, \dots, b_{\frac{p-1}{2}}=\zeta^{\frac{p-1}{2}}+\zeta^{-\frac{p-1}{2}}$. Then the coefficient matrix is formed by rows formed with 0 and -2, (0 or 1 more -2). $\endgroup$– Angel del RioCommented Apr 7, 2015 at 6:57
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